Science:Math Exam Resources/Courses/MATH152/April 2010/Question B 05 (b)/Solution 1

Continuing from the hint, to choose real-valued solutions, we choose $C_2=\overline{C_1}$, so that the general solution becomes:

$\mathbf{𝐲}=C_1{\mathbf{y}}_{\mathbf{𝟏}}+C_2{\mathbf{y}}_{\mathbf{𝟐}}=C_1{\mathbf{y}}_{\mathbf{𝟏}}+\overline{C_1{\mathbf{y}}_{\mathbf{𝟏}}}=2\mathrm{R}\mathrm{e}\left[C_1{\mathbf{y}}_{\mathbf{𝟏}}\right]$

Recall that

${\mathbf{y}}_{\mathbf{𝟏}}={\mathbf{k}}_{\mathbf{𝟏}}\exp {\lambda }_{1}t=\left[\begin{array}{c}2+3i\\ 1\end{array}\right]\exp (1+2i)t$

Since

$\mathrm{R}\mathrm{e}\left[{\mathbf{y}}_{\mathbf{𝟏}}\right]=\left[\begin{array}{c}2\\ 1\end{array}\right]e^t\cos 2t+\left[\begin{array}{c}-3\\ 0\end{array}\right]e^t\sin 2t$

$\mathrm{I}\mathrm{m}\left[{\mathbf{y}}_{\mathbf{𝟏}}\right]=\left[\begin{array}{c}3\\ 0\end{array}\right]e^t\cos 2t+\left[\begin{array}{c}2\\ 1\end{array}\right]e^t\sin 2t$

The general form of real-valued solutions is:

$\begin{array}{rl}\mathbf{𝐲}& =D_1\mathrm{R}\mathrm{e}\left[{\mathbf{y}}_{\mathbf{𝟏}}\right]+D_2\mathrm{I}\mathrm{m}\left[{\mathbf{y}}_{\mathbf{𝟏}}\right]\\ & =D_1(\left[\begin{array}{c}2\\ 1\end{array}\right]e^t\cos 2t+\left[\begin{array}{c}-3\\ 0\end{array}\right]e^t\sin 2t)\\ & +D_2(\left[\begin{array}{c}3\\ 0\end{array}\right]e^t\cos 2t+\left[\begin{array}{c}2\\ 1\end{array}\right]e^t\sin 2t)\end{array}$

where $D_1$ and $D_2$ are real constants.

Their relationship with the previous constants are $D_1=2\mathrm{R}\mathrm{e}\left[C_1\right]$ and $D_2=-2\mathrm{I}\mathrm{m}\left[C_1\right]=e^t$

Now if we impose the initial condition $\mathbf{𝐲}(0)=\left[\begin{array}{c}1\\ 0\end{array}\right]$, we get

$D_1\left[\begin{array}{c}2\\ 1\end{array}\right]+D_2\left[\begin{array}{c}3\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]$

Thus, $D_1=0$ and $D_2=\frac{1}{3}$, and the solution of the initial value problem is

$\frac{1}{3}(\left[\begin{array}{c}3\\ 0\end{array}\right]e^t\cos 2t+\left[\begin{array}{c}2\\ 1\end{array}\right]e^t\sin 2t)=\frac{e^t}{3}\left[\begin{array}{c}3\cos 2t+2\sin 2t\\ \sin 2t\end{array}\right]$

Alternatively, one can solve directly from

$\mathbf{𝐲}=2\mathrm{R}\mathrm{e}\left[C_1{\mathbf{y}}_{\mathbf{𝟏}}\right]$

And impose the initial condition ${\mathbf{𝐲}}_0=\left[\begin{array}{c}1\\ 0\end{array}\right]$ direcly to find that $C_1=-\frac{i}{6}$.

Then compute the real part to get the same result as above. The amount of algebra involved will be slightly less.